$\{\sup_n X_n \leq a\} = \cap_n \{X_n \leq a\}$

Question

I would like to understand why this statement is true:

$\{\sup_n X_n \leq a\} = \cap_n \{X_n \leq a\}$

Could I have an example or a justification? Here $X_n$ is a random variable.

This will also help me understand the set-theoretic definitions of limsup and liminf.

Thank you!

Answer 1

It follows from the following:

Let $X_n:\Omega\rightarrow\mathbb{R}$ be a sequence of functions and let $a\in\mathbb{R}$. Then $\sup_{n\in\mathbb{N}} X_n(\omega)\leq a$ if and only if $X_n(\omega)\leq a$ for all $n\in\mathbb{N}$.

Proof: If $\sup_{n\in\mathbb{N}} X_n(\omega)\leq a$ then clearly all $X_n(\omega)\leq a$. For the other direction assume by contradiction that $\sup_{n\in\mathbb{N}} X_n(\omega)=b>a$ choose $\varepsilon<b-a$ then by the definition of supremum there exists $n$ such that $X_n(\omega)>b-\varepsilon>a$ which is absurd.

• Note that the statement in the question is not completely mathematical. A more accurate version would be $$\{\omega \in \Omega : \sup_{n\in\mathbb{N}} X_n(\omega)\leq a\} = \bigcap_n \{\omega \in \Omega : X_n(\omega)\leq a\}$$

Answer 2

In short:$$\omega\in\{\sup_n X_n\leq a\}\iff\sup_nX_n(\omega)\leq a\iff\forall n\;X_n(\omega)\leq a\iff\omega\in\bigcap_{n=1}^{\infty}\{X_n\leq a\}$$

hence:$$\{\sup_n X_n\leq a\}=\bigcap_{n=1}^{\infty}\{X_n\leq a\}$$